A question about ideals, maximal ideals, and UFD's.

Question

I'm working on the following problem in preparation for an exam.

Let $f(x) = x^3 + 2 \in \mathbb{Z}[x]$. Let $\alpha$ be any complex number satisfying $f(x)$. Let $K = \mathbb{Q}(\alpha)$ and suppose $R$ is the collection of element in $K$ with integer coefficients (with basis $1, \alpha$, and $\alpha^2$).

Next we are asked to show the following two equalities:

1. For every non-zero ideal $I \subseteq \mathbb{Z}$, show that $IR \cap \mathbb{Z} = I$.
2. Show that if $J \subseteq R$ is a nonzero ideal , then $J \cap \mathbb{Z} \neq \emptyset$.
3. Show that every nonzero prime ideal in $R$ is maximal.

Here is my work so far: It is clear that $IR$ contains $I$ and $\mathbb{Z}$ contains $I$. Thus $IR \cap \mathbb{Z} \supseteq I$. Now suppose $x \in IR \cap \mathbb{Z}$. Then $x = ir$ for some $i \in I$ and $r \in R$. Also, $ir \in \mathbb{Z}$. Now I am not sure how to proceed. I want to show that $x \in I$.

I know that $\mathbb{Z}$ is a UFD, and I believe that saying $x$ has a unique factorization in $\mathbb{Z}$ should help. In fact, since $x \in \mathbb{Z}$ and $I \subseteq \mathbb{Z}$ then $ir \in \mathbb{Z}$ says that $r$ is also in $\mathbb{Z}$. But then $x$ is a multiple of an element in $I$, and so $x \in I$. (is this correct?)

Unless I am mistaken, $0 \in J$ for every nonempty $J\subseteq R$.

I'm not sure where to start on part 3. I know what prime ideals are, and what maximal ideals are.

If someone has a better suggestion for the title, please let me know. I wasn't sure how to paraphrase this question.

Answer 1

So you have seen that $I \subset IR \cap \mathbb Z$, and are confused about the other way.

The point is, note that every ideal of $\mathbb Z$ is of the form $n \mathbb Z = \{nz : z \in \Bbb Z\}$. So, write $I = n\Bbb Z$.

Now, let $x \in IR \cap \mathbb Z$. Clearly, $x$ is an integer. Furthermore, note that $x = ir$ for some $i \in I, r \in R$. We know what $I,R$ are : this helps us conclude that $i = jn$ for some integer $j$ and $r = d + e\alpha + f \alpha^2$ for integers $d,e,f$.

Thus, $x = jnd + jne\alpha + jnf\alpha^2$. Transposing $x$ to the other side, $jnf \alpha^2 + jne\alpha + jnd - x = 0$. So $\alpha$ satisfies a polynomial of degree smaller than $3$ with integer coefficients. This polynomial must be a factor of $x^3 +2$, but that is irreducible, so this cannot happen. Consequently, the above polynomial must be zero. That is, all the coefficients are zero : $jnf = 0$,$jne = 0$ and ,more importantly, $jnd - x = 0$ or $x = jnd$, that is $x \in I$. Hence, the first part is complete.

---

For the second part, we know $0 \in J \cap \mathbb Z$, but let us make it stronger. Let $p(x)$ be a non-zero polynomial such that $p(\alpha) \in J$. Since $\alpha^3 = 2$, we can assume $1 \leq \deg p \leq 2$ (if $\deg p = 0$ then $p$ is a non-zero constant, so we are done). Now, using the division algorithm, write $x^3 + 2 = p(x)q(x) + r(x)$, where $r(x)$ has degree less than $p(x)$, and is non-zero because $x^3 + 2$ is irreducible. Substituting $\alpha$, we get $0 = p(\alpha) q(\alpha) + r(\alpha)$. Thus, $r(\alpha) = p(\alpha) \times -q(\alpha) \in J$.

Now, $\deg p = 1 \implies \deg r = 0$ so $J$ contains a non-zero constant. Now, if $\deg p = 2$ then either $\deg r = 0$,whence we are done, or $\deg r = 1$, whence we replace $p$ by $r$ in the paragraph above to get the conclusion. Either way, $J$ contains a non-zero constant, so $J \cap \mathbb Z$ is a set which contains more elements than just zero. In fact, let $n$ be the smallest non-zero positive integer contained in $J \cap \Bbb Z$. Of course, for any $m$, we have $nm \in J \cap \Bbb Z$. However, if using division algorithm , $m = nq + r \in J \cap \mathbb Z$, then $r \in J \cap \mathbb Z$ and $r < n$, contradiction. Consequently, $J \cap \mathbb Z = n\mathbb Z$ with $n$ as above.

---

Let $J$ be a prime ideal of $R$. Then, $\frac RJ$ is an integral domain. Now, note that $J \cap \mathbb Z = n \mathbb Z$, so therefore, $(n \mathbb Z) R \subset JR = J$. Therefore, it also follows that $\frac RJ \subset \frac R{(n \mathbb Z) R}$ (the latter is the quotient of $R$ by the ideal generated by the constant polynomial $n$ in $R$).

Now, define the map $R \to \left(\frac{\mathbb Z}{n\mathbb Z}\right)^3$ by $a\alpha^2 + b\alpha+c \to ([a],[b],[c])$, where $[x]$ denotes remainder of $x$ when divided by $n$. Clearly, $(n \mathbb Z)R$ is the kernel of this map, so it follows that the cardinality of $\frac R{(n \mathbb Z)}$ is finite, since it is less than or equal to $n^3$ (infact, it is equal, you can see surjectivity of the map easily). Finally, it follows that $\frac RJ$ is a finite integral domain, and these are known to be fields (the popular pigeonhole argument). Consequently, $\frac RJ$ is a field, so $J$ is maximal.

Answer 2

For what it's worth, I have an alternate approach.

1.) Considering that $\mathbb Z$ is a PID and $I$ is a nonzero ideal of $\mathbb Z,$ there exists a nonzero integer $n$ such that $I = n \mathbb Z.$ Consequently, every element of $IR \cap \mathbb Z$ is an integer $r$ of the form $r = nk(a + b \alpha + c \alpha^2),$ where $a, b, c,$ and $k$ are also integers. By viewing $R$ as a $\mathbb Q$-vector space with a $\mathbb Q$-basis of $\{1, \alpha, \alpha^2\},$ it follows by the uniqueness of a vector in $R$ that $b = c = 0,$ and we conclude that $r = n(ka)$ is in $I = n \mathbb Z.$ QED.

2.) Clearly, the intersection of a nonzero ideal $J$ of $R$ and $\mathbb Z$ is nonempty: by definition, $J$ contains $0,$ hence we have that $0 \mathbb Z \subseteq J \cap \mathbb Z.$ Unfortunately, there was a typo in the question; the actual problem statement asks us to prove that every nonzero ideal $J$ of $R$ satisfies $J \cap \mathbb Z \neq 0 \mathbb Z.$ Consider a nonzero ideal $J$ of $R.$ Observe that $R \cong \mathbb Z[x] / (x^3 + 2),$ hence $R$ is Noetherian. Consequently, the ideal $J$ of $R$ has a primary decomposition $J = Q_1 \cap \cdots \cap Q_n$ by nonzero primary ideals $Q_i$ such that $\sqrt{Q_i} = P_i$ is a nonzero prime ideal. Considering that radicals distribute over intersections, we have that $\sqrt J = P_1 \cap \cdots \cap P_n.$ By definition of the radical, we have that $(\sqrt{J})^k \subseteq J$ for some integer $k \gg 0,$ hence we have that $$J \supseteq (P_1 \cap \cdots \cap P_n)^k \supseteq (P_1 \cdots P_n)^k = P_1^k \cdots P_n^k.$$ Considering that $R$ is an integral extension of $\mathbb Z$ (because $1, \alpha,$ and $\alpha^2$ are integral over $\mathbb Z$), it follows that $\dim R = \dim \mathbb Z = 1,$ hence every nonzero prime ideal of $R$ is maximal. Contracting any nonzero prime ideal $P_i$ of $R$ to $\mathbb Z$ gives a nonzero maximal ideal $P_i \cap \mathbb Z = p_i \mathbb Z$ for some prime $p_i.$ Ultimately, we find that $J \cap \mathbb Z \supseteq P_1^k \cdots P_n^k \cap \mathbb Z \supseteq (P_1 \cap \mathbb Z)^k \cdots (P_n \cap \mathbb Z)^k = p_1^k \cdots p_n^k \mathbb Z.$ QED.

3.) Once we know that every nonzero ideal $J$ of $R$ satisfies $J \cap \mathbb Z \supseteq 0 \mathbb Z,$ it follows that $J \cap \mathbb Z = n \mathbb Z$ for some nonzero integer $n.$ Given that $J$ is a prime ideal of $R,$ we must have that one of the prime factors of $n$ is an element of $J.$ Call that prime factor $p.$ Observe that the action $(k + p \mathbb Z) \cdot (r + J) = kr + J$ is well-defined: indeed, if we have that $k + p \mathbb Z = \ell + p \mathbb Z,$ then $k - \ell$ is in $p \mathbb Z$ so that $k - \ell$ is in $J$ and $kr + J = \ell r + J.$ Consequently, we may view $R / J$ as a $\mathbb Z / p \mathbb Z$-module. But as $\mathbb Z / p \mathbb Z$ is a field, the integral domain $R / J$ is a finite-dimensional $\mathbb Z / p \mathbb Z$-vector space and hence must be a field. We conclude that $J$ is maximal. QED.